# [SOLVED] Bounds of Integration

• Apr 23rd 2010, 04:51 PM
[SOLVED] Bounds of Integration
I have two problems that are pretty similar:

1. Use a triple integral to find the volume of the solid bound by the parabolic cylinder $\displaystyle y=9x^2$ and the planes $\displaystyle z = 0, z = 9$, and $\displaystyle y = 5$.

I have been having issues finding the x bounds on this integral, but this is what I have tried: I set $\displaystyle y = 5 = 9x^2$ and solved for x, which gave me $\displaystyle x = + \sqrt {\frac {5} {9}}$ and $\displaystyle x = - \sqrt {\frac {5} {9}}$

$\displaystyle \int_{- \sqrt {\frac {5} {9}}}^{\sqrt {\frac {5} {9}}}\int_{9x^2}^{5}\int_{0}^{9} dzdydx$

I was just curious if this is right or not since I have been trouble with figuring out the bounds.

2.

The second question is this:

Find the volume of the solid enclosed by the paraboloids $\displaystyle z = (x^2 +y^2)$ and $\displaystyle z = 32 - (x^2+y^2)$

The bounds for z are there, but to find them for y and x could I set z = 0 and solve? So that $\displaystyle z = (x^2 + y^2)$ gives bounds for x and y that are 0 and then upper bounds for x and y that are $\displaystyle \sqrt {32}$.

And so the integral would be:

$\displaystyle \int_{0}^{\sqrt{32}}\int_{0}^{\sqrt{32}}\int_{(x^2 + y^2)}^{32 - (x^2 + y^2)} dzdydx$

Thanks ahead of time for the help.
• Apr 23rd 2010, 07:33 PM
TKHunny
I'm going to agree with the first one, but include a small chastisement. Please exploit symmetry where you can. Multiply the whole thing by 2 and change the first lower limit to zero (0).

On the second, I have to object. Why do you think the extremes would be at z = 0? Frankly, z = 16 seems more likely. Further, why would you not change this one to Polar Coordinates?
• Apr 23rd 2010, 10:24 PM
I am sorry and I don't mean to be a pain, but could you work out the second one? I haven't learned about converting a triple integral into polar coordinates, but I would appreciate the direction.
• Apr 24th 2010, 04:50 AM
TKHunny
Why is "triple" any different than "double", as far as this is concerned? It would be converting only the x-y components. This would make it Cyllindrical Coordinates.

$\displaystyle x = r\cdot \cos(\theta)$
$\displaystyle y = r\cdot \sin(\theta)$

Implication: $\displaystyle r^{2} = x^{2}+y^{2}$

Implication: $\displaystyle dx\;dy\;=\;r\;dr\;d\theta$

The very first side effect you should notice is the ease at which you can now find the appropriate limits.

Do NOT forget to exploit symmetries.
• Apr 24th 2010, 03:28 PM
Quote:

Originally Posted by TKHunny
Why is "triple" any different than "double", as far as this is concerned? It would be converting only the x-y components. This would make it Cyllindrical Coordinates.

$\displaystyle x = r\cdot \cos(\theta)$
$\displaystyle y = r\cdot \sin(\theta)$

Implication: $\displaystyle r^{2} = x^{2}+y^{2}$

Implication: $\displaystyle dx\;dy\;=\;r\;dr\;d\theta$

The very first side effect you should notice is the ease at which you can now find the appropriate limits.

Do NOT forget to exploit symmetries.

Ah, thank you. That was much easier. I appreciate it a lot, thank you again.